NECO Mathematics 2002 past question

NECO Mathematics 2002 · try it first

What are the factors of 2x

^{2}

- xy - 3y

^{2}

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To factor $2 x^{2} - x y - 3 y^{2}$ assume $(a x + b y) (c x + d y)$ with $a c = 2$ , $b d = - 3$ and $a d + b c = - 1$ .
Take $a = 2$ , $c = 1$ (since $2 \cdot 1 = 2$ ). Then $b d = - 3$ gives either $(b, d) = (3, - 1)$ or $(- 3, 1)$ .
Check $a d + b c$ :

For $(b, d) = (3, - 1)$ , $a d + b c = 2 (- 1) + 3 = 1 \neq = - 1$ .
For $(b, d) = (- 3, 1)$ , $a d + b c = 2 (1) + (- 3) = - 1$ , which satisfies the condition.

Thus the factorisation is $(2 x - 3 y) (x + y)$ , which matches option D.

The most tempting wrong choice is C, $(2 x + 3 y) (x - y)$ , because it also yields $2 x^{2}$ and $- 3 y^{2}$ but gives $+ x y$ instead of $- x y$ .

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